Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $t \neq 0$. $n = \dfrac{3(2t - 9)}{2} \times \dfrac{t}{2t - 9} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ 3(2t - 9) \times t } { 2 \times (2t - 9) } $ $ n = \dfrac {t \times 3(2t - 9)} {2 (2t - 9)} $ $ n = \dfrac{3t(2t - 9)}{2(2t - 9)} $ We can cancel the $2t - 9$ so long as $2t - 9 \neq 0$ Therefore $t \neq \dfrac{9}{2}$ $n = \dfrac{3t \cancel{(2t - 9})}{2 \cancel{(2t - 9)}} = \dfrac{3t}{2} $